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Let $S \subset C([0,1],\mathbb{R})$ be an $\mathbb{R}$-linear subspace that is invariant under the $T := \int_0^x$ integration operation: if $g \in S$ then the function $f = Tg$ defined pointwise by $f(x) := \int_0^x g(t) \, dt$ is in $S$ too.

A basic example is the (unital) polynomial algebra $A := \mathbb{R}[x] \hookrightarrow C([0,1],\mathbb{R})$, which according to Weierstrass's theorem is dense in the uniform (i.e., $L^{\infty}$) norm. That means that in the $L^{\infty}$ norm, the closure $\overline{A} = C([0,1],\mathbb{R})$. Generalizing this, consider more generally the closure $\overline{S}$ of $S$ in the $L^{\infty}$ norm. The manifest possibilities for that closure are the linear subspaces $$ \{f \in C([0,1],\mathbb{R}) \mid f|_{I} \equiv 0 \} $$ defined by a vanishing condition along some open ($I = [0,a)$) or closed ($I = [0,a]$) initial segment $0 \in I \subset [0,1]$, possibly empty or reduced to the point $\{0\}$.

Question. Are there any other possibilities for the closure $\overline{S}$ besides these?

Equivalently, and in the contrapositive formulation: If $g \in C([0,1],\mathbb{R})$ has $g(0) \neq 0$, must the constant function $1 = \chi_{[0,1]}$ be a uniform limit of linear combinations from $\{T^ng \mid n = 0,1,\ldots\}$?

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  • $\begingroup$ Possible idea: show that your contrapositive formulation is true if $g$ is a polynomial, then use that a general $g$ is the uniform limit of polynomials $p_m$. Since linear combinations of $\{T^n p_m\}$ are dense for any $m$, and since $T^n p_m$ is close to $T^n g$ for $m$ large (depending on $n$), you can try to use a diagonalization argument to show some linear combination of $T^n g$ is close to $1$. $\endgroup$
    – user101142
    Commented Apr 6, 2018 at 15:09
  • $\begingroup$ @user37208: I am not sure that such a diagonal approximation argument would work, for what if say our function $g \not\equiv 0$ vanishes identically on $[0,0.5]$? Then $g$ is still a uniform limit of polynomials, but of course all $T^n g$ vanish along the initial segment $[0,0.5]$ (and so do all uniform limits of their finite linear combinations). With polynomials, contrastingly, the contrapositive statement does always hold with either $I = \emptyset$ or $I = \{0\}$. $\endgroup$
    – Vesselin Dimitrov
    Commented Apr 6, 2018 at 15:18
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    $\begingroup$ I believe it is a known theorem that all closed invariant subspaces in $C[0,1]$ for this operator are of the obvious form (functions vanishing on $[0,a]$ for some $a\ge 0$) but I wouldn't stake my head on that, just suggest that you do a literature search with key words like "volterra operator" "invariant subspace", etc. $\endgroup$
    – fedja
    Commented Apr 6, 2018 at 15:48
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    $\begingroup$ @fedja: Thank you, that does it: it turns out a classical theorem due independently to S. Agmon (Sur une probleme de translations, C.R. Acad. Sci Paris, 1949) and, in a generalized form, G.K. Kalish (On similarity, reducing manifolds, and unitary equivalence of certain Volterra operators, Ann. Math., 1957). $\endgroup$
    – Vesselin Dimitrov
    Commented Apr 6, 2018 at 16:14

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